Question

Write the value of ${{\cot }^{-1}}\left( -x \right)$ for all $x\in R$ in terms of ${{\cot }^{-1}}x$.

Answer 1

Hint: In this given question, we must first of all convert the given equation, that is ${{\cot }^{-1}}\left( -x \right)$ in terms of $\cot $. Thereafter we can use the property of cotangent function to express x as a cotangent of some angle. Then, we can again take ${{\cot }^{-1}}$ on both sides and use the given relation to obtain the required answer.

Complete step-by-step answer:

In this given question, we are first of all going to convert the given equation, that is ${{\cot }^{-1}}\left( -x \right)$ in terms of $\cot $as follows:
Let, ${{\cot }^{-1}}\left( -x \right)=\theta $………………….(1.1)
$\Rightarrow \left( -x \right)=\cot \theta $
$\Rightarrow x=-\cot \theta ........................(1.2)$
Now, we know that when we increase the value of an angle by a multiple of $\pi $, the value of its cotangent becomes negative, thus we can write
$\cot (n\pi +\theta )=-\cot \left( \theta \right)...............(1.3)$
where n is an integer.
Using this in equation (1.2), we get
$x=\cot \left( n\pi -\theta \right)$
$\Rightarrow {{\cot }^{-1}}\left( x \right)=n\pi -\theta $………………………. (1.4)
Now, from equation 1.1, we get the value of $\theta $ as equal to ${{\cot }^{-1}}\left( -x \right)$. So, putting this value in equation 1.4, we get,
${{\cot }^{-1}}\left( x \right)=n\pi -\theta $
$\Rightarrow {{\cot }^{-1}}\left( x \right)=n\pi -{{\cot }^{-1}}\left( -x \right)$
$\Rightarrow {{\cot }^{-1}}\left( -x \right)=n\pi -{{\cot }^{-1}}\left( x \right)$……………… (1.5)
Hence, from equation 1.5, we get the required expression we wanted to prove.
Therefore, we have expressed the value of ${{\cot }^{-1}}\left( -x \right)$ for all $x\in R$ in terms of ${{\cot }^{-1}}x$.

Note: We should note that in equation (1.2), we could have written a more general expression for x as
$x=-\cot \left( \theta +r\pi \right)$ where r in an integer because the value of cot remains the same if we increase the value of the angle by a multiple of $\pi $. However, in this case, we would have got equation (1.4) as ${{\cot }^{-1}}\left( x \right)=n\pi -\left( \theta +r\pi \right)$ which is equivalent to $n\pi -{{\cot }^{-1}}\left( x \right)$. Thus, in both the methods, the final answer will remain the same.